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MODELING AND ANALYSIS OF MANUFACTURING SYSTEMS Session 12 MACHINE SETUP AND OPERATION SEQUENCING E. Gutierrez-Miravete Spring 2001. INTRODUCTION. WHOLE SYSTEM DESIGN STRATEGIES VS INDIVIDUAL CELL/WORKSTATION DESIGN STRATEGIES - PowerPoint PPT PresentationTRANSCRIPT
MODELING AND ANALYSIS OFMANUFACTURING SYSTEMS
Session 12 MACHINE SETUP AND
OPERATION SEQUENCING E. Gutierrez-Miravete
Spring 2001
INTRODUCTION• WHOLE SYSTEM DESIGN
STRATEGIES VS INDIVIDUAL CELL/WORKSTATION DESIGN STRATEGIES
• QUESTION: HOW TO TOOL THE MACHINE AND THEN SEQUENCE PRODUCTION ACTIVITIES IN IT?
• GOAL: MAXIMIZE PRODUCTIVITY
OPORTUNITIES
• SEQUENCE BATCHES SO AS TO MINIMIZE TOOLING CHANGEOVERS
• SEQUENCE ACTIVITIES SO AS TO MINIMIZE IDLE TIMES
• OPTIMIZE CELL LAYOUT SO AS TO MINIMIZE ASSEMBLY TIME
SEQUENCING AND OPTIMIZATION
MANY SEQUENCING PROBLEMS IN MANUFACTURING CELL PLANNING ARE CLASSIC OPTIMIZATION PROBLEMS
CELL VS SYSTEM• BOTH, CLEVER SEQUENCING OF
OPERATIONS AND OVERALL SYSTEM DESIGN ARE IMPORTANT FOR SUCCESS IN MANUFACTURING
• TIME FRAMES– FOR OVERALL SYSTEM DESIGN: WEEKS
OR MONTHS (LONG TERM)– FOR SEQUENCE DESIGN:
MINUTES/HOURS (SHORT TERM)
TASK ASSIGNMENT
LINEAR ASSIGNMENT PROBLEM
• GOAL:– TO DISTRIBUTE N TASKS
AMONG N WORKERS SO AS TO MINIMIZE COST
• CONSTRAINTS:– 1 TASK PER WORKER– 1 WORKER PER TASK
COST MATRIX C
• ROWS: WORKERS• COLUMNS: MACHINES• SUMMARIZES THE
ASSIGNMENT COSTS
LAP: MATHEMATICAL FORMULATION
• minimizeCi j cij xij
• subject to:i xij = 1 (for all tasks)
j xij = 1 (for all workers)
Facts
• IF A CONSTANT IS ADDED TO EVERY ELEMENT OF A ROW OR COLUMN OF C, THE OPTIMAL SOLUTION DOES NOT CHANGE BUT ITS VALUE CHANGES BY THE ADDED CONSTANT
• IF ALL cij > 0, ANY SOLUTION WITH COST = 0 MUST BE OPTIMAL
Hungarian Algorithm
• Proceeds by adding and substracting constants from rows and columns so as to maintan a non-negative cost matrix. When a feasible solution is found using only 0 cost cells, optimum has been found.
HA Steps
• STEP 1: COST REDUCTION BY CONSTRUCTION OF THE REDUCED COST MATRIX. THE RCM IS OBTAINED BY SUBSTRACTING FIRST THE MINIMUM ELEMENT IN EACH ROW FROM ALL ELEMENTS IN THE ROW THEN DOING LIKEWISE WITH COLUMNS
HA Steps cont’d
• STEP 2: SEARCH FOR A FEASIBLE SOLUTION USING ONLY THE 0’S IN THE RCM . IF THIS SUCCEEDS, OPTIMAL SOLUTION HAS BEEN FOUND. IF ALL 0’S CAN BE COVERED WITH LESS THAN n HORIZONTAL AND VERTICAL LINES, CONTINUE.
HA Steps cont’d
• STEP 3: FURTHER REDUCTION. FIND THE MINIMUM UNCOVERED ELEMENT. SUBSTRACT THIS REDUCED COST FROM EACH UNCOVERED ELEMENT AND ADD IT TO EACH TWICE-COVERED ELEMENT. GO TO 2.
• Example 8.1, Table 8.1, Fig. 8.1, pp. 263-
TASK SEQUENCING
TWO CLASSES OF PROBLEMS
• EVALUATE CHANGEOVER COSTS WHEN COSTS ARE COMPLETELY DETERMINED BY THE CURRENT JOB SETUP AND THE NEXT JOB TO BE LOADED
• EVALUATE COSTS WHEN THE ENTIRE SEQUENCE OF JOBS MUST BE KNOWN
COMPLETE CHANGEOVERS• N JOBS ARE TO BE PERFORMED ON
ONE MACHINE• UNIT PROCESSING TIME AND BATCH
SIZE DETERMINE TOTAL PROCESSING TIME
• CHANGEOVER TIMES DEPEND ONLY ON CURRENT AND NEXT PRODUCT
• TOTAL SETUP TIME DEPENDS ON JOB SEQUENCE
TRAVELING SALESMAN PROBLEM
• A SALESMAN MUST VISIT EVERY CITY IN HIS/HER TERRITORY THEN RETURN HOME IN SUCH A WAY THAT THE SMALLEST POSSIBLE TOTAL DISTANCE IS TRAVELLED.
• TSP CAN BE VISUALIZED WITH A GRAPH OF NODES (CITIES) AND ARC LENGTHS (DISTANCES)
TRAVELING SALESMAN PROBLEM
• DISTANCE BETWEEN CITIES i & j = cij
• ALL ARCS ARE BIDIRECTIONAL• NO SUBTOURS ALLOWED• FIVE CITY COMPLETE TSP GRAPH
Fig 8.2a• FIVE CITY POSSIBLE TOUR GRAPH
Fig 8.2b
SOLUTION OF THE TSP• CLASSICAL OPTIMIZATION
TECHNIQUES ARE HARD TO APPLY WHEN N IS LARGE (See Eqn 8.5-)
• HEURISTIC METHODS ARE MOST FREQUENTLY USED– NO SUBTOUR CONSTRAINT IS
TEMPORARILY RELAXED– SOLVE RESULTING OPTIMIZATION
PROBLEM
TSP BY CLOSEST INSERTION ALGORITHM
1.- SELECT A STARTING CITY2.- PROCEED THROUGH N-1 STAGES
ADDING A NEW CITY AT EACH STAGE. THE NEW CITY IS SELECTED FROM THOSE CURRENTLY UNASSIGNED SUCH THAT IT IS CLOSEST TO ANY CITY IN THE ACCUMULATED PARTIAL SEQUENCE
• Example 8.2; Fig 8.3, pp. 268-
TSP BY MINIMUM SPANNING TREE
• A MST IS ANY SET OF N-1 ARCS THAT TOUCH EACH NODE AND HAVE THE SMALLEST SUM OF COSTS FOR ANY SUCH SET
• BY MODIFYING THE MST SO THAT EACH NODE IS CONNECTED EXACTLY BY TWO ARCS IN A CONNECTED TREE OBTAIN TSP SOLUTION
TSP BY SUBTOUR INTEGRATION
• START WITH A SOLUTION TO THE ASSIGNMENT PROBLEM
• TRY TO CONNECT TWO SUBTOURS AT A TIME BY SWITCHING ARCS
• ONCE ALL LAP SUBTOURS ARE COMBINED, GOT TSP SOLUTION
• Fig 8.4
PARTIAL CHANGEOVERS• N JOBS ARE TO BE PERFORMED ON ONE
MACHINE CAPABLE OF HOLDING M TOOLS• JOB j REQUIRES TOOLS Aj• TOTAL NUMBER OF TOOLS REQUIRED
EXCEEDS M (TOOL CHANGES REQUIRED)• NO JOB REQUIRES MORE THAN M TOOLS• AT EACH JOB COMPLETION SOME TOOLS
MAY HAVE TO BE REMOVED AND NEW TOOLS ADDED
• Fig 8.5
OBJECTIVE
TO ORDER JOBS AND TOOL CHANGEOVERS TO MINIMIZE THE TOTAL NUMBER OF TOOLS CHANGED ON THE MACHINE
NOTES
• ALWAYS KEEP M TOOLS ON MACHINE
• THE ORDERING FOR JOBS ON THE MACHINE IS GIVEN
• KEEP TOOL NEEDED SOONEST (KTNS) RULE IS OPTIMAL
• REMOVE ONLY “LONGEST UNTIL NEXT USE” TOOLS
JOB ORDERING
• IF JOB r USES ONLY A SUBSET OF TOOLS USED BY ITS PREDECESOR s CHANGEOVER IS NOT INCREASED (i.e. OPTIMAL SEQUENCE)
• JOB SEQUENCING PROBLEM IS SIMILAR TO GROUP FINDING IN GT
• THE TOOL-JOB MATRIX
JOB ORDERING
• SOLUTION METHODS– BINARY CLUSTERING
• Ex 8.3, Tables 8.3, 8.4
– TSP• Table 8.5
SOLUTION OF THE PARTIAL CHANGEOVER
PROBLEM
• STEP 1: JOB COMBINATION (REDUCTION)
• STEP 2: JOB ORDERING• STEP 3: TOOL SETUP PLANNING
BY KTNS
INTEGRATED ASSIGNMENT AND SEQUENCING
QUESTION
• WHAT TO DO WHEN CELL SETUP (TOOLING) AND JOB SEQUENCING ARE RELATED BY NEITHER DICTATES THE OTHER?
• NEED TO:– MAKE A SEQUENCING DECISION and– MAKE A SETUP DECISION
NOTES
• PROBLEM 1: PLAN SETUP AND OPERATION OF ASSEMBLY CELLS
• PROBLEM 2: SETUP AND SEQUENCE OF A MACHINE (INTERDEPENDENT TOOLS)
• GOAL: TO MINIMIZE CYCLE TIME• SOLUTION TECHNIQUES:
HEURISTICS
ASSEMBLY CELL LAYOUT AND SEQUENCING
• ASSEMBLY ENVIRONMENTS– MASS PRODUCED SINGLE PRODUCT– MULTIPLE PRODUCTS PRODUCED IN
ALTERNATING LOTS WITH CHANGEOVER REQUIRED
– MIX OF PART TYPES ASSEMBLED SIMULTANEOUSLY IN CELL WITHOUT CHANGEOVER
REQUIRED
• FIND HOW TO PERFORM THE SET OF TASKS THAT HAVE BEEN ASSIGNED TO A WORKSTATION
SINGLE PART TYPE
• A SINGLE PRODUCT FRAME• PRODUCT FRAME HAS N
LOCATIONS WHERE PARTS ARE TO BE ADDED
• WORKSTATION HAS N BINS WHERE PART FEEDERS ARE PLACED
• Fig 8.6
REQUIRED
• ASSIGN FEEDERS TO BINS• DETERMINE THE ORDER IN
WHICH PARTS ARE TO BE ADDED TO FRAME
• PROBLEM CAN BE MODELED AS A 2N-CITY TSP
BIN ASSIGNMENT AND INSERTION SEQUENCING
• BIN ASSIGNMENT GOAL: MINIMIZE THE LOADED TRAVEL TIME (DISTANCE) FOR THE ASSEMBLER
• INSERTION SEQUENCING GOAL: MINIMIZE THE UNLOADED TRAVEL TIME (DISTANCE) FOR THE ASSEMBLER
• Ex 8.4, Tables 8.6 , 8.7; Ex 8.5, Table 8.8
MIXED PRODUCTS
• TYPICALLY, SEVERAL PRODUCT TYPES ARE BEING PRODUCED SIMULTANEOUSLY IN THE SAME CELL
• DEMAND PROPORTIONS DETERMINED BY BILL OF MATERIALS FOR END PRODUCTS
MIXED PRODUCT: UNPACED LINE
• ASSUME – RELATIVE DEMANDS ARE KNOWN– M FRAMES TO BE MADE– pm IS THE PROPORTION OF TYPE m
FRAMES– BINS KEPT ON SAME PLACE
• GOAL: MINIMIZE AVERAGE ASSEMBLY TIME OF PRODUCT
MIXED PRODUCT: UNPACED LINE
• FOR EACH PART TYPE COMPILE TABLE OF TRAVEL TIMES (Table 8.7)
• COMBINE TABLES BY WEIGHTED AVERAGE
cij = m pm cijm
– pm proportion of type m frames– cijm total travel time per type m frame if part i
is assigned to bin j– Ex. 8.6, Tables 8.9, 8.10
CELL LAYOUT AND SEQUENCING:
INTERDEPENDENT TOOLS• THE NC PUNCH PRESS
– 36 TOOL TOOL TURRET (Fig 8.7)– PARTS MAY REQUIRE UP TO 200 HITS– HIT SEQUENCE MAY BE SUBJECT TO
PRECEDENCE CONSTRAINTS• GOAL: LOAD TOOLS AND SEQUENCE
HITS TO MINIMIZE PRODUCTION CYCLE• Ex 8.7, Fig. 8.8, Tables 8.11, 8.12